Problem: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $46.4$ years; the standard deviation is $9.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living between $55.8$ and $74.6$ years.
Explanation: $46.4$ $37$ $55.8$ $27.6$ $65.2$ $18.2$ $74.6$ $99.7\%$ $68\%$ $15.85\%$ $15.85\%$ We know the lifespans are normally distributed with an average lifespan of $46.4$ years. We know the standard deviation is $9.4$ years, so one standard deviation below the mean is $37$ years and one standard deviation above the mean is $55.8$ years. Two standard deviations below the mean is $27.6$ years and two standard deviations above the mean is $65.2$ years. Three standard deviations below the mean is $18.2$ years and three standard deviations above the mean is $74.6$ years. We are interested in the probability of a bear living between $55.8$ and $74.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the bears will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $68\%$ of the bears will have lifespans within 1 standard deviation of the mean. The probability of a particular bear living between $55.8$ and $74.6$ years is $\color{orange}{15.85\%}$.